Be $X$ and $Y$ two compact Hausdorff spaces, we will say that an application $T: C (X; \mathbb{R}) \rightarrow C (Y; \mathbb{R})$ is positive if for every function $f \ge 0$ we have that $T (f) \ge 0$.
$C (X; \mathbb{R})$: functions defined on $X$, a compact Hausdorff space, which are continuous and bounded, with norm $||f||_\infty=\sup_{x\in X}|f(x)|$
Show that if the application $T: C (X; \mathbb{R}) \rightarrow C (Y; \mathbb{R})$ is a positive linear application, then $T$ is a continuous application.
My ideas:
$T: C (X; \mathbb{R}) \rightarrow C (Y; \mathbb{R})$
$\qquad\qquad f\longrightarrow T(f)=T(f(y))$
P.D:
$\exists M>0$ such that $||T(f)||_\infty \le M||f||_\infty$
Now, $||T(f)||_\infty=||T(\frac{f}{f})f||_\infty=||T(1)f||_\infty=||T(1)||_\infty ||f||_\infty$, where $1$ is the function constant defined by $1:X\rightarrow \mathbb{R}$ s.t $1(x)=1$
$||T(f)||_\infty=||T(1)||_\infty||f||_\infty$, let $M=||T(1)||_\infty>0$. Finally, $||T(f)||_\infty= M||f||_\infty$.
Therefore, T is bounded $\Rightarrow$ T is continuous.
Is correct my proof?
And other question:
The other part of the problem is to prove that $||T(1)||_{\infty}=||T||$ where $1$ is the function constant defined by $1:X\rightarrow \mathbb{R}$ s.t $1(x)=1$
$||T(1)||_{\infty}\le ||T||||1||_{\infty}=||T||\Rightarrow ||T(1)||_{\infty}\le||T||$
The other side: $||T||\le||T(1)||_{\infty}$
Let $||f||_{\infty}\le 1$, with $f\ge 0,\forall x\in X$.
Now $|f(x)|\le||f||_{\infty}\le 1\Rightarrow 1\ge |f(x)|=f(x),\forall x\in X$
$\Rightarrow f(x)\le 1\Rightarrow 0\le 1-f(x)\Rightarrow 0\le 1-f $
With $T$ positive we have that $0\le T(1-f)=T(1)-T(f)\Rightarrow T(f)\le T(1)$
Therefore, $\sup_{x\in X}|T(f)(x)|\le \sup_{x\in X}|T(1)(x)|\Rightarrow ||T(f)||_{\infty}\le||T(1)||_{\infty}$.
Finally,
$||T||=\sup_{||f||_{\infty}=1}||T(f)||_{\infty}\le||T(1)||_{\infty}\Rightarrow ||T||\le ||T(1)||_{\infty}$.
We conclude that $||T||=||T(1)||_{\infty}$
Since positive elements span $C(X;\mathbb R)$, you need only check that $T$ is bounded on the set of positive functions. (For if $\|Tf\|\leq M\|f\|$ for $f\geq 0$, then for arbitrary $f$ we have $\|Tf\|=\|Tf^+-Tf^-\|\leq M\|f^+\|+M\|f^-\|\leq2M\|f\|$.)
Assume the opposite. Then you obtain a sequence of positive functions $\{f_n\}$ in $C(X;\mathbb R)$ such that $\|f_n\|\leq1$ while $\|Tf_n\|\geq4^n$. Then $f=\sum_n2^{-n}f_n$ is positive, and $f\geq2^{-n}f_n$ for all $n$, so $Tf\geq 2^{-n}Tf_n$ and thus
$$\|Tf\|\geq2^{-n}\|Tf_n\|\geq2^n$$ for all $n\in\mathbb N$, a contradiction.